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emmainna [20.7K]
3 years ago
7

Enter the missing numbers in the boxes to complete the table of equivalent ratios of time to distance.

Mathematics
2 answers:
mylen [45]3 years ago
6 0
The ratio between 12 and 8 is 1.5, so you need to divide every other number by 1.5.


2/1.5=1.3

9/1.5=6

18/1.5=12

If I read, and understood, your question correctly, this should be right.
ozzi3 years ago
4 0

Answer:

For Time (h) the order is

3

9

12

18

For Distance (mi) the order is

2

6

8

12

All together it looks like:

3:2

9:6

12:8

18:12

Step-by-step explanation:

I took the test online in K12 and when I reviewed it said that that is the correct numbers and order. Have a nice day :)

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4 is your answer

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Find the length of the third side. If necessary, write in simplest radical form.
Angelina_Jolie [31]

Answer:

the length of the missing side is 2

Step-by-step explanation:

Use a^2 + b^2 = c^2

a = 4

b = ?

c = 2sqrt5

4^2 + b^2 = (2sqrt5)^2

16 + b^2 = 20

subtract 16 from both sides

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3 0
3 years ago
At a recent marathon, spectators lined the street near the starting line to cheer for the runners. The crowd lined up 5 feet dee
masha68 [24]

Answer:

29568 people cheered for the runners at the start of the race

Step-by-step explanation:

From the question, the crowd lined up 5 feet deep on both sides of the street for the first mile.

This lined up crowd could be related to a rectangle that is 1 mile long and 5 feet wide.

First, Convert 1 mile to feet

1 mile = 5280 feet

Hence, the length of the rectangle is 5280 feet and the width is 5 feet.

Now, we will determine how many 5 feet by 5 feet square we can get from the 5280 feet by 5 feet rectangle. To do that, we will divide 5280 feet by 5 feet

5280 feet ÷ 5 feet = 1056

Hence, from the rectangle, we can get 1056 5 feet by 5 feet square.

From the question, you estimate that 14 people can comfortably fit in a square that measures 5 feet by 5 feet,

∴ 14 × 1056 people will comfortably fit in the crowed lined up 5 feet deep on one side of the street for the first mile.

14 × 1056 = 14784 people

This is the amount of people that will comfortably fit in the crowed lined up 5 feet deep on one side of the street for the first mile.

Since the crowd lined up on both sides of the street, then

2 × 14784 people will comfortably fit in the crowed lined up 5 feet deep on both sides of the street for the first mile

2 × 14784 = 29568 people

Hence, 29568 people cheered for the runners at the start of the race.

5 0
4 years ago
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